Form of coprime ideals in commutative rings

Question

I was trying to prove that there exists a bijection between the idempotent elements in a commutative ring $R$, and possible factorisations of $R$ into two other commutative rings $R_{1} \times R_{2}$. With not too much difficulty, it can be shown that for every idempotent element $i$ there is a factorisation, the converse is pretty hard though. I've reduced the problem to the question if for every pair of coprime ideals $I, J$ in $R$ there exists an idempotent $i \in R$ such that $I = iR$ and $J = (1-i)R$. Is this statement true? Or should I start my proof over again? If it is true, could I get a good hint on how to prove that statement?

Answer 1

The product $R_1 \times R_2$ has two obvious idempotents, namely $(1,0)$ and $(0,1)$, and indeed, $R_1 \times R_2 = (R_1 \times R_2)(1,0) \times (R_1 \times R_2) (0,1)$. Letting $\phi: R_1 \times R_2 \overset{\sim}{\to} R$ be the given isomorphism, we simply push this factorization forward by $\phi$.

Let $e = \phi(1,0)$. Note that $\phi(1,1) = 1$ (since $(1,1)$ is the multiplicative identity in the product $R_1 \times R_2$), so $$ \phi(0,1) = \phi((1,1) - (1,0)) = \phi(1,1) - \phi(1,0) = 1 - e $$ as desired.

Answer 2

I would simply, for a factorisation $R=R_1\times R_2$, consider $e=p_{1}(1)$ and show that $e$ is an idempotent of $R$, $1-e$ is the orthogonal idempotent, and that $R_1=Re$, $R_2=R(1-e)$.